Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e5 + 5;
const int lg = 17;
int n, m, x, y, z, q, par[17][maxn], h[maxn];
vector<int> g[maxn];
vector<int> s[lg][maxn];
void dfs(long long v, long long p = -1) {
  for (auto u : g[v])
    if (u != p) h[u] = h[v] + 1, par[0][u] = v, dfs(u, v);
}
int find(int v, int x) {
  for (int i = 0; i < lg; i++)
    if ((x) & (1 << i)) v = par[i][v];
  return v;
}
int LCA(int x, int y) {
  if (h[x] < h[y]) swap(x, y);
  x = find(x, h[x] - h[y]);
  for (int i = lg - 1; i >= 0; i--)
    if (par[i][x] != par[i][y]) x = par[i][x], y = par[i][y];
  if (x == y)
    return x;
  else
    return par[0][x];
}
void merge(vector<int> &s, vector<int> t, vector<int> p) {
  int l = 0, r = 0;
  vector<int> v;
  for (int i = 0; i < 10; i++) {
    if (t.size() <= l and p.size() <= r) break;
    if (t.size() > l and (r >= p.size() or t[l] <= p[r]))
      v.push_back(t[l]), l++;
    else
      v.push_back(p[r]), r++;
  }
  s = v;
}
vector<int> solve(int v, int l, bool f = 0) {
  int x = h[v] - h[l] + 1 - f;
  vector<int> t;
  for (int i = lg; i >= 0; i--)
    if ((1 << i) & x) {
      merge(t, t, s[i][v]);
      v = par[i][v];
    }
  return t;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m >> q;
  for (int i = 1; i < n; i++) {
    cin >> x >> y, x--, y--;
    g[x].push_back(y), g[y].push_back(x);
  }
  dfs(0);
  for (int i = 0; i < m; i++) cin >> x, x--, s[0][x].push_back(i);
  for (int i = 0; i < n; i++) sort(s[0][i].begin(), s[0][i].end());
  for (int j = 1; j < lg; j++)
    for (int i = 0; i < n; i++) {
      par[j][i] = par[j - 1][par[j - 1][i]];
      merge(s[j][i], s[j - 1][i], s[j - 1][par[j - 1][i]]);
    }
  for (int h = 0; h < q; h++) {
    int l, a, u, v;
    vector<int> ans, ans1, ans2;
    cin >> v >> u >> a, v--, u--;
    l = LCA(v, u);
    if (l == v)
      ans1 = solve(u, l);
    else if (l == u)
      ans2 = solve(v, l);
    else {
      ans1 = solve(u, l);
      ans2 = solve(v, l, 1);
    }
    merge(ans, ans1, ans2);
    z = min((int)ans.size(), a);
    cout << z << " ";
    for (int i = 0; i < z; i++) cout << ans[i] + 1 << " ";
    cout << endl;
  }
  return 0;
}